This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
let\'s do a direct translation from hexadecimal to binary like this:
E79A.6A416 = E 7 9 A. 6 A 4 = E(=1110) 7(=0111) 9(=1001) A(=1010). 6(=0110) A(=1010) 4(=0100) = 1110011110011010.01101010012
answer: E79A.6A416 = 1110011110011010.01101010012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
14∙163+7∙162+9∙161+10∙160+6∙16-1+10∙16-2+4∙16-3 = 14∙4096+7∙256+9∙16+10∙1+6∙0.0625+10∙0.00390625+4∙0.000244140625 = 57344+1792+144+10+0.375+0.0390625+0.0009765625 = 59290.415039062510
got It: E79A.6A416 =59290.415039062510
Translate the number 59290.415039062510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
59290 | 2 | | | | | | | | | | | | | | | |
-59290 | 29645 | 2 | | | | | | | | | | | | | | |
0 | -29644 | 14822 | 2 | | | | | | | | | | | | | |
| 1 | -14822 | 7411 | 2 | | | | | | | | | | | | |
| | 0 | -7410 | 3705 | 2 | | | | | | | | | | | |
| | | 1 | -3704 | 1852 | 2 | | | | | | | | | | |
| | | | 1 | -1852 | 926 | 2 | | | | | | | | | |
| | | | | 0 | -926 | 463 | 2 | | | | | | | | |
| | | | | | 0 | -462 | 231 | 2 | | | | | | | |
| | | | | | | 1 | -230 | 115 | 2 | | | | | | |
| | | | | | | | 1 | -114 | 57 | 2 | | | | | |
| | | | | | | | | 1 | -56 | 28 | 2 | | | | |
| | | | | | | | | | 1 | -28 | 14 | 2 | | | |
| | | | | | | | | | | 0 | -14 | 7 | 2 | | |
| | | | | | | | | | | | 0 | -6 | 3 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 1 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 4150390625*2 |
0 | .83008*2 |
1 | .66016*2 |
1 | .32031*2 |
0 | .64063*2 |
1 | .28125*2 |
0 | .5625*2 |
1 | .125*2 |
0 | .25*2 |
0 | .5*2 |
1 | .0*2 |
the result of the conversion was:
59290.415039062510 = 1110011110011010.01101010012
answer: E79A.6A416 = 1110011110011010.01101010012